Capacitive Filters

Discussion in 'Homework Help' started by Jaziek, Apr 21, 2011.

  1. Jaziek

    Thread Starter New Member

    Apr 21, 2011
    4
    0
    I have a high pass and low pass filter, which look like this,

    [​IMG]

    R1 = 1kΩ R2 = 10kΩ , and C1 = 10 nF

    I need to find the breakpoint values for each of these filters. When I put them into LTSpice, it gives me 14.1kHz for the low pass filter, and 1.8kHz for the high pass filter.

    But when I use the equation 1/(2∏RC) to work out what the values should be ideally, it gives me something noticeably different.

    I know that this is because I'm not properly taking into account the second resistor, but I don't know how I to do this.

    How do I work out cutoff frequencies for filters in circuits with more than 1 resistor?

    thanks :p
     
    Last edited: Apr 21, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you add a load resistor Rload, then for the purpose of analysis R2 is replaced with Rload||R2. [ || meaning "in parallel with" ]
     
  3. Jaziek

    Thread Starter New Member

    Apr 21, 2011
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    Sorry I guess I phrased my question badly, used terminology wrongly and whatnot, I'm very new to electronics in general.

    What I mean is, I don't know how to properly account for R2, because I've only ever seen examples of filters with 1 resistor.
     
  4. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    Your question demonstrates the inadequacy of merely memorizing "formulas." Looking at case B, notice that we are seeking the voltage across R2, which is part of a basic voltage divider. Therefore:

    v_o=\frac{R_2}{R_2+R_1+\frac{1}{j\omega C_1}}v_i

    Dividing through by R2+R1, we get:

    v_o=(\frac{R_2}{R_2+R_1})\frac{1}{1-j\frac{1}{\omega C_1(R_2+R_1)}}v_i

    Now we have separated the resistive component of the divider, which attenuates the signal regardless of frequency, from the reactive component, which IS dependent on frequency (note the presence of ω). Half power (cutoff) occurs when:

    \frac{1}{sqrt{1-j\frac{1}{\omega C_1(R_2+R_1)}^2}}=\frac{1}{\sqrt{2}}

    Therefore, at cutoff:

    \omega C_1(R_2+R_1)=1

    Solving for ω:

    \omega=\frac{1}{C_1(R_2+R_1)}

    Convert angular frequency to cyclical and you have your answer. This technique (NOT this formula) can be applied to part A as well.
     
    Last edited: Apr 21, 2011
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  5. Jaziek

    Thread Starter New Member

    Apr 21, 2011
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    It does indeed, but everybody has to start somewhere :p

    Thanks for helping. I'll give this a try.
     
  6. Jaziek

    Thread Starter New Member

    Apr 21, 2011
    4
    0
    so if I understand this right, which I probably dont,

    For the low pass filter, I need to find the parallel impedance of R2 and C1, like this...

     Z_p = \frac {Z_1 Z_2} {Z_1 + Z_2}

    where z1 and z2 are r2 and c1

    right?

    then put that into a voltage divider with R1, which gives

     \frac {V_o}{V_i} = \frac {Z_p}{R_1 + Z_p}

    I'm having trouble simplifying what I get at this point to anything meaningful though.
     
  7. Heavydoody

    Active Member

    Jul 31, 2009
    140
    11
    Actually, yes you do. Good work.

    Yes, it gets a bit tricky here. Try dividing through with Zp. Then expand Zp and continue simplifying from there. You should eventually end up with:

    \frac{v_o}{v_i}=(1+\frac{R_1}{R_2})\frac{1}{1+j \frac{\omega R_1R_2C_1}{R_2+R_1}}
     
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